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Gauß famously determined the cubic character of $2$ in his Disquisitiones : $2$ is a cube modulo a prime number $p\equiv1\mod3$ if and only if $p=x^2+27y^2$ for some $x,y\in\mathbf{Z}$. This implies that the prime numbers which split completely in the $\mathfrak{S}_3$-extension $\mathbf{Q}(\root3\of1,\root3\of2)$ of $\mathbf{Q}$ are precisely the ones which are $\equiv1\pmod3$ and represented by the quadratic form $X^2+27Y^2$.

This was generalised by Philippe Satgé in 1977 in a paper whose title I have borrowed. He shows for example that the prime numbers which split completely in the $\mathfrak{S}_3$-extension $\mathbf{Q}(\root3\of1,\root3\of5)$ of $\mathbf{Q}$ are precisely the ones which are $\equiv1\pmod3$ and represented by one of the quadratic forms $$ X^2+XY+169Y^2,\qquad 343X^2-131XY+13Y^2. $$ I believe that similar results about all $\mathfrak{S}_3$-extensions (of $\mathbf{Q}$) can now be recovered using the known cases of Langlands reciprocity, as illustrated around the same time by Serre for the splitting fields of $T^3-T-1$ and $T^3+T-1$, which are the maximal unramified abelian extensions of $\mathbf{Q}(\sqrt{-23})$ and $\mathbf{Q}(\sqrt{-31})$ respectively, in his Modular forms of weight one and Galois representations, pp. 193–268 of Algebraic number fields: L-functions and Galois properties (Proc. Sympos., Univ. Durham, Durham, 1975), Academic Press, London, 1977.

But Satgé's theorem is applicable to more general extensions : it is applicable to a $G$-extension $K$ of $\mathbf{Q}$ whenever the finite group $G$ contains a commutative normal subgroup $H\subset G$ such that

(*) the transfer (Verlagerung) map $G/G'\to H$ is trivial, and

(**) the order of $H$ is odd if the field $K^H$ is totally real of degree $>2$ (over $\mathbf{Q}$).

Question. Can these more general results of Satgé be recovered from known cases of Langlands reciprocity ?

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Chandan -- the reason one can recover these quadratic forms from Langlands is: (1) $S_3$ has a faithful 2-d rep (2) 2-d reps with finite solvable image are known to be modular by Langlands-Tunnell and (3) if the corresponding repn is odd and the image is dihedral then the appropriate weight 1 modular form can be built from theta series. But if any of these things fail you're probably in trouble. But I am a little skeptical about your statement of Satg\'e's theorem. Does it really apply to $G$ a general non-abelian finite simple group and $H$ the trivial subgroup? ... –  user30035 Jan 19 '13 at 16:28
... If so then you're safe from the Langlands people, because non-soluble base change is still a long way off from these Langlands people; they can say nothing other than conjectures. –  user30035 Jan 19 '13 at 16:29
@wccanard: Yes, your remarks are what I had in mind when I said that the decomposition laws for $\mathfrak{S}_3$-extensions can be recovered from Langlands. And yes, Satgé's theorem does seem to apply to a finite simple group $G$ (take $H$ to be trivial), but then the split primes are no longer characterised by quadratic forms but by forms of degree equal to the order of $G$. –  Chandan Singh Dalawat Jan 20 '13 at 3:55
... forms of degree $|G/H|$ in $|G/H|$ variables. –  Chandan Singh Dalawat Jan 20 '13 at 5:21
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1 Answer

Far from being able to even begin to answer the question, let me make a few remarks that perhaps shed some light on the situation.

Gauss's result on the cubic character of $2$ was generalized in Dedekind's highly underrated article on pure cubic fields [Über die Anzahl der Idealklassen in reinen kubischen Zahlkörpern; J. Reine Angew. Math. 121 (1900), 40-123]: already in the early 1870s Dedekind had written

Let $k$ denote a rational integer whose cube root is irrational. Then the equation $x^3 = k$ defines a pure cubic number field whose discriminant has the form $D = -3g^2$, where the integer $g$ can easily be determined from $k$. Consider all primes $p$ of the form $3n+1$ that do not divide $k$ and modulo which the given integer $k$ is a cubic residue. With the help of the reciprocity law we then find the following interesting result, which essentially was known already to Gauss (and may be extended to general cubic fields): there are three kinds of primitive binary quadratic forms $ax^2 + bxy + cy^2$ with $D = b^2 - 4ac$ and representing different classes: the forms of the first kind are a group whose forms represent exactly those prime numbers modulo which $k$ is cubic residue.

He could not generalize his results, as promised, to general cubics, although he correctly conjectured that this should be possible using the theory of complex multiplication, i.e., class field theory for complex quadratic number fields.

Dedekind also quotes a remark by Gauss, in which he determined the cubic character of $5$ using binary quadratic forms (which coincides with Satgé's example except that Gauss considered only forms with even middle coefficient, which means that he has four forms $(1,0,675)$, $(25,0,27)$, $(13,2,52)$, $(4,2,169)$ where Satgé can do with one $(1,1,169)$.

Dedekind's result was generalized by Takagi [Sur les corps résolubles algébriquement, C. R. Acad. Sci. Paris 171 (1920), 1202-1205]. Takagi presents Dedekind's theorem in the following form:

Let $D$ denote the discriminant of a cubic number field $k$; then the class number of primitive quadratic forms with discriminant $D$ is a multiple of $3$. A third of the classes forms a group which can be characterized by the property that the prime numbers not dividing $D$ and for which $D$ is a quadratic residue, and which split into three different factors in $k$, and only those, are represented by a quadratic form in this group.

Then he proves his generalization:

Let $k$ be a solvable field of prime degree and $K_0$ the corresponding cyclic field. If the ideal classes in $K_0$ are defined modulo $f$, the class group contains a subgroup of index $\ell$, which is characterized by the property that among the prime numbers not dividing the discriminant of $k$ and splitting completely in $K_0$, exactly those primes that split completely in $k$ are the norms of ideals in $K_0$ lying in the subgroup above.

If $k$ is a pure cubic field, then $K_0$ is the field of cube roots of unity, and Takagi gets back Dedekind's result. For the proof, Takagi used his class field theory.

Special cases of this result were rediscovered e.g. by

  • D. Liu [Dihedral polynomial congruences and binary quadratic forms in her Ph.D. thesis (Carleton 1992) supervised by
  • Spearman & Williams [The cubic congruence $x^3 + Ax^2 + Bx + C \equiv 0 \pmod p$ and binary quadratic forms, J. London Math. Soc. (2) 46 (1992), no. 3, 397-410], [The cubic congruence $x^3 + Ax^2 + Bx + C \equiv 0 \pmod p$ and binary quadratic forms II, J. London Math. Soc. (2) 64 (2001), no. 2, 273-274].

See also

  • D. Bernardi [ Résidus de puissances, Semin. Delange-Pisot-Poitou 1977/78, Fasc. 2, Exp. No. 28, 12 pp.

The results by

  • Weinberger [The cubic character of quadratic units, Proc. 1972 Number Theory Conf., Univ. Colorado, Boulder 1972, 241-242]

and the more recent ones by

  • Sun [Cubic residues and binary quadratic forms, J. Number Theory (2006)]

also follow this pattern (there are in fact a lot more articles dealing with describing the splitting of primes in ring class fields of quadratic number fields using binary quadratic forms).

Satgé considers the following situation: let $K$ be a normal extension of the rationals with Galois group $G$, let $H$ be a normal abelian subgroup, and let $k$ be the fixed field of $H$. By applying class field theory to the abelian extension $K/k$, he characterizes the decomposition law in $K$ by representation of the primes in question by the norm forms attached to $k$, assuming certain conditions on $H$. The norm forms have degree $(G:H)$; thus the case where $H = 1$ is absolutely trivial since in this case the splitting of primes in $K$ (for unramified primes in normal extensions, all we need to know is the inertia degree) is described by norms from $K$.

Overall these are all "abelian" phenomena and follow readily from (abelian) class field theory. But of course it is legitimate to ask how these fit into the nonabelian Langlands conjectures, since after all we are dealing with nonabelian extensions here.

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Many thanks for such a wonderfully informative answer, Franz. –  Chandan Singh Dalawat Feb 7 '13 at 2:58
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